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In plain GR, geodesic are defined by:

$$ \nabla_{u} u^{\,\mu} = 0 $$

where $u^{\,\mu}$ is the four-velocity of the particle.

Now, I don't understand if this holds true also in the electromagnetic case, or if in this case the geodesics equation should be

$$ \nabla_{u} u^{\,\mu} = \frac{q}{m} F^{\mu \nu}u_{\,\nu} $$

with $q$ the particle charge, $m$ is the particle mass and $F^{\mu \nu}$ the Faraday tensor.

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  • $\begingroup$ I believe the second formula should have an $m$ (for mass) rather than an $n$ $\endgroup$ Commented May 30 at 2:19
  • $\begingroup$ How many are there? One geodesic equation? Several geodesic equations? Should it a geodesic equation in the first sentence? Or the geodesic equation? Or geodesic equations? Or the geodesic equations? Or a geodesic curve? Or geodesic curves? Or the geodesic curve? Or the geodesic curves? Or something else? $\endgroup$ Commented May 30 at 19:20

3 Answers 3

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Geodesics are still the same: they are the paths that a particle in free fall follows.

However, charged particles are no longer in free fall. They are subject to the Lorentz force. This is described by prescribing to them an acceleration given in terms of the Faraday tensor $F_{ab}$, which is your second formula. Notice that the charge and the mass occur in that equation as well.

In summary, charged particles don't move on geodesics, because they are not in free fall.

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  • $\begingroup$ Does this mean that electrons can never move on geodesics since they are charged and massive? Not even in an electrovacuum? $\endgroup$
    – Aleph12345
    Commented May 30 at 3:11
  • $\begingroup$ @AndreaDiPinto That is correct. Electrons do not move in geodesics. $\endgroup$ Commented May 30 at 14:53
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    $\begingroup$ I should mention that there are also corrections to both the geodesic motion of neutral particles and to the non-geodesic motion of charged particles. In first approximation, those equations yield good results, but there are corrections due to the interaction of the particle with its own gravitational or electromagnetic fields. This is discussed in the last chapter of Wald's Advanced Classical Electromagnetism, which references his papers with Gralla on the subject $\endgroup$ Commented May 30 at 14:55
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    $\begingroup$ So technically, no particle actually moves in geodesics. But those are very good approximations. $\endgroup$ Commented May 30 at 14:55
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In presence of external forces the motion is non-geodesic, as explained in other answers. One way to see it is that the full action of a point particle in a curved background that is coupled to electromagnetism is given by

$$ S = \int(m-qA_\mu \frac{dx^\mu}{d\lambda})\, d \tau$$

where $d \tau = \sqrt{-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}} d \lambda$. Stationarity of the action will give you the equivalent of your second equation. The equation of motion is valid, but it is not considered "geodesic", as the particle is not in free fall and a non-gravitational interaction is considered.

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No, the geodesic equation is the analog of $x''=0$ (N2L with no external force). With the presence of electromagnetism, the equations of motion become as you have stated and is the analog of $x''=F/m$ with a nonzero (electromagnetic) force. Now, while this is the equation of motion, it is not the geodesic equation. The geodesic equation is what a particle would do under the influence of no external force, that is it (locally) moves in a straight line. The geodesic equation is purely geometric, while the force law is more physical. With respect to the geometry, you can measure the acceleration of the particle moving, hence the "error in the geodesic" or "force" in the right side of the equation. You can think of the EM force as analogous to an external force of an SHO or other such classical system.

Now, you can "couple" to the Levi civita connection (i.e. add) the electromagnetic potential, which is mathematically a connection on an auxiliary $U(1)$ i.e. $\mathbb{C}$ bundle (technically, this is a connection on $TM\otimes L$ with $L$ the line bundle chosen, which is usually trivial). Taking the geodesic equation w.r.t. this coupled connection gives the force law of EM which you wrote down. So in a sense, it is a geodesic equation, but not the purely geometric (i.e. from the manifold itself) geodesic. This sorta corresponds to saying that if you keep twisting your local frame to counteract the electromagnetic force, you won't see any acceleration. Big words, but the summary is "kinda". More generally, these tricks give Yang Mills equations by coupling the Levi civita connection with other "gauge connections".

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  • $\begingroup$ It doesn’t really make sense to speak of a geodesic equation on $TM\otimes L$, so the “kinda” should really be be “no”. Geodesics (or autoparallel curves) are properties of a connection (here the Levi-Civita connection) on the tangent bundle of a manifold alone. $\endgroup$
    – peek-a-boo
    Commented May 30 at 2:54
  • $\begingroup$ I may be mixing things up, but can you not just apply the coupled covariant derivative twice and set it to 0, and then the derivatives of A give the curvature? $\endgroup$ Commented May 30 at 3:01
  • $\begingroup$ Oh I suppose that is working with fields, it doesn't work for particles. $\endgroup$ Commented May 30 at 3:11
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    $\begingroup$ The key property that allows us to absorb gravity into the geometry is the universality of gravitational motion: the motion does not depend on the mass (or charge, or any other property of the object). This means that geodesics are defined only by the particle's initial position and velocity. This is not the case for electromagnetism. $\endgroup$
    – Rd Basha
    Commented May 30 at 3:15
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    $\begingroup$ $E=TM\otimes L$ is a vector bundle over $M$, so yes given connections on the two factors we do get a natural connection on the tensor product. So for a curve $\gamma$ in $M$, we can definitely make sense of $\nabla^E_{\dot{\gamma}}$. However, this cannot be applied to $\dot{\gamma}$ (i.e yes, it cannot be applied to a particle’s velocity), because $\dot{\gamma}$ is a curve in $TM$, and this cannot naturally be interpreted as a curve in $E=TM\otimes L$. So, bottom line is only $\nabla_{\dot{\gamma}}\dot{\gamma}$ can be made sense of, but $\nabla^E_{\dot{\gamma}}\dot{\gamma}$ can’t. $\endgroup$
    – peek-a-boo
    Commented May 30 at 3:23

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